Question

When concentration of fch2cooh (ka = 2.6 ×10*-3 ) is needed so that (h+) = 2 × 10*-3?

Answer 1

Ka = 2.57 e - 3  

2.6 e - 3 = [H+] [FCH2COO-] / [FCH2COOH]  

2.6 e - 3 = 2.0 e - 3 ( 2 e -3) / [FCH2COOH]  

[FCH2COOH] = 1.53 e -3 when at equilibrium  

Initial conc. of acid = concentration at equilibrium + concentration of H+ produced  

1.53 e - 3 + 2.0 e - 3 = 3.53 e - 3 M

Answer 2

The concentration of $FCH_2COOH needed so that [tex][H^+]=2\times 10^{-3}$ is 0.004 M

Explanation:

$FCH_2COOH\rightleftharpoons H^+FCH_2COO^-$

    cM              0             0

$c-c\alpha$        $c\alpha$          $c\alpha$  

So dissociation constant will be:

$K_a=\frac{(c\alpha)^{2}}{c-c\alpha}$

Given : $K_a=2.6\times 10^{-3}$ and $[H^+]=2\times 10^{-3}$

$[H^+]=c\alpha$

$c\alpha=2\times 10^{-3}$

Putting in the values we get:

$2.6\times 10^{-3}=\frac{(2\times 10^{-3})^2}{(c-2\times 10^{-3})}$

$c=0.004M$

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